Interpreting 1 H nmr spectra

1. Interpreting 1H nmr spectra L.O.: Intrepet 1H nmr spectra using the n+1 rule

2. H - C - X - C - H CHEMICAL SHIFT Low d ‘shielding’ Bonding to electronegative atoms (O, N) High d ‘deshielding’ Approximate chemical shifts The actual values depend on the environment ROH -CHO -COOH -C=CH- TMS 13 12 11 10 9 8 7 6 5 4 3 2 1 0 d ‘deshielding’

3. LOWRESOLUTION • low resolution NMR gives 1 peak for each environmentally different group of protons • Strengths of the absorption are proportional to number of equivalent 1H atoms. It is measured by the are under each peak. Integration. LOW RESOLUTION SPECTRUM OF 1-BROMOPROPANE

4. Look at low resolution 1H NMR of EtOH

5. The simplified NMR spectrum of EtOH shows three single peaks. A detailed, high-resolution spectrum of EtOH shows that some peaks are split into a number of subsidiary peaks. This splitting is caused by spin-spin coupling between protons on neighbouring atoms.

6. High resolution 1H NMR of EtOH.

7. The ‘n +1’ rule The n.m.r. absorption of a proton which has n equivalent neighbouring protons will be split into n + 1 peaks.

9. MULTIPLICITY (Spin-spin splitting) O adjacent H’s There is no effect 1 adjacent H can be aligned either with a or against b the field there are only two equally probable possibilities the signal is split into 2 peaks of equal intensity

10. MULTIPLICITY (Spin-spin splitting) O adjacent H’s There is no effect 1 adjacent H can be aligned either with a or against b the field there are only two equally probable possibilities the signal is split into 2 peaks of equal intensity 2 adjacent H’s more possible combinations get 3 peaks in the ratio 1 : 2 : 1

11. MULTIPLICITY (Spin-spin splitting) O adjacent H’s There is no effect 1 adjacent H can be aligned either with a or against b the field there are only two equally probable possibilities the signal is split into 2 peaks of equal intensity 2 adjacent H’s more possible combinations get 3 peaks in the ratio 1 : 2 : 1 3 adjacent H’s even more possible combinations get 4 peaks in the ratio 1 : 3 : 3 : 1

12. MULTIPLICITY (Spin-spin splitting) Number of peaks = number of chemically different H’s on adjacent atoms + 1 1 neighbouring H 2 peaks “doublet” 1:1 2 neighbouring H’s 3 peaks “triplet” 1:2:1 3 neighbouring H’s 4 peaks “quartet” 1:3:3:1 4 neighbouring H’s 5 peaks “quintet” 1:4:6:4:1 Signals for the H in an O-H bond are unaffected by hydrogens on adjacent atoms - get a singlet

13. INTEGRATION • the area under a signal is proportional to the number of hydrogen atoms present • an integration device scans the area under the peaks • lines on the spectrum show the relative abundance of each hydrogen type • By measuring the distances between the integration lines one can • work out the simple ratio between the various types of hydrogen. • before integration after integration NOTICE THAT THE O-H SIGNAL IS ONLY A SINGLET

14. INTEGRATION Measure the distance between the top and bottom lines. Compare the heights from each signal and make them into a simple ratio. • HOW TO WORK OUT THE SIMPLE RATIOS • Measure how much each integration line rises as it goes of a set of signals • Compare the relative values and work out the simple ratio between them • In the above spectrum the rises are in the ratio... 1:2:3 IMPORTANT: It doesn’t provide the actual number of H’s in each environment, just the ratio

15. 1 2 3 4 NMR SPECTROSCOPY When is a hydrogen chemically different? TWO SIGNALS Quartet and triplet :- ratio of peak areas = 3 : 2 Carbons 1 & 4 are the similar and so are carbons 2 & 3 so there are only two different chemical environments. The signal for H’s on carbon 2 is a quartet - you ignore the two neighbours on carbon 3 because they are chemically identical. BUTANE

16. 1 2 3 4 NMR SPECTROSCOPY When is a hydrogen chemically different? TWO SIGNALS Quartet and triplet :- ratio of peak areas = 3 : 2 Carbons 1 & 4 are the similar and so are carbons 2 & 3 so there are only two different chemical environments. The signal for H’s on carbon 2 is a quartet - you ignore the two neighbours on carbon 3 because they are chemically identical. TWO SIGNALS both singlets :- ratio of peak areas = 2 : 1 Hydrogens on OH groups only give singlets. The signal for H’s on each carbon are not split, because - H’s on the neighbouring carbon are chemically identical... and - H’s on adjacent OH groups do not couple. BUTANE ETHANE-1,2-DIOL

17. NMR SPECTROSCOPY - SUMMARY An nmr spectrum provides several types of information :- number of signal groups tells youthe number of different proton environments chemical shift the general environment of the protons peak area (integration) the number of protons in each environment multiplicity how many protons are on adjacent atoms In many cases this information is sufficient to deduce the structure of an organic molecule but other forms of spectroscopy are used in conjunction with nmr.

18. NMR spectra of –OH and –NH protons • They are usually broad • The is usually no splitting pattern.

19. D2O Shake CH3CH2OH + D2O → CH3CH2OD + HOD

21. 1H NMR TASK 4 For each of the following compounds, draw the molecule predict the number of signals predict the relative intensity of each signalpredict the approximate chemical shift (of each signal a)propanoic acid b)propanal c)2-chloropropane d)2-methylbutane e)methylpropene f)methyl propanoate

24. Task 5-9

25. Interpreting 1H nmr spectra L.O.: Intrepet 1H nmr spectra using the n+1 rule

26. WHAT IS IT! C2H5Br 3 2

27. WHAT IS IT! C2H3Br3 2 1

28. WHAT IS IT! C2H4Br2 3 1

29. WHAT IS IT! C2H4O2 1 3 1

30. WHAT IS IT! C4H8O2 3 3 2

31. WHAT IS IT! C3H6O

32. WHAT IS IT! C3H6O 3 2 1

33. WHAT IS IT! C4H8O 3 3 2

34. WHAT IS IT! C8H16O2

35. WHAT IS IT! C11H16

36. WHAT IS IT! C8H10

37. WHAT IS IT! C8H10 3 2

38. WHAT IS IT! C9H12

39. WHAT IS IT! C4H8Br2